Prove that the set of functions $\sin kx$ is closed and bounded

Question

Given a set of functions $M=\lbrace f_k \rbrace$, where $f_k = \sin (kx)$, prove that M is closed and bounded.

I have been given a norm: $$\| f\| = \left[ \int_{0}^{2 \pi} f (x)^2 dx \right]^{1/2}$$ (on the space of continuous, real functions from $[0,2\pi]$ to $\mathbb{R}$), but no metric or topology. I don't see how I can do this.

Answer 1

General remarks

• A norm induces a metric: $d(x,y) = \|x-y\|$.
• A metric induces a topology, via neighborhoods $N(x,r) = \{y:d(x,y)<r\}$. Which becomes $N(x,r) = \{y:\|x-y\|<r\}$ when the metric comes from a norm.

Specific set

Boundedness amounts to a direct computation: $\|f_k\| = \pi^{1/2}$

The reason the set is closed is that its points are uniformly far apart; as a result, the set has no limit points. To justify this, compute $\|f_k-f_j\| = (2\pi)^{1/2}$ for $k\ne j$.

The computations are based on double-angle and product-to-sum identities.